16.5 Translating Words into Algebraic Symbols 2In order to solve application problems, it is necessary to translate English phrases into algebraic symbols. The following are some common phrases and their mathematic translation. 3 Applications

Translating from Words to Mathematical Expressions

Mathematical Expression (where x and y are numbers)Verbal ExpressionAdditionThe sum of a number and 2x 23 more than a numberx 37 plus a number7 x16 added to a numberx 16A number increased by 9x 9The sum of two numbersx y 4 Applications

Translating from Words to Mathematical Expressions

Mathematical Expression (where x and y are numbers)Verbal ExpressionSubtraction4 less than a numberx 410 minus a number10 xA number decreased by 5x 5A number subtracted from 1212 xThe difference between two numbersx y 5 Applications

Translating from Words to Mathematical Expressions

Mathematical Expression (where x and y are numbers)Verbal ExpressionMultiplication14 times a number14xA number multiplied by 88x3xTriple (three times) a numberxyThe product of two numbers 6 Applications

Translating from Words to Mathematical Expressions

Mathematical Expression (where x and y are numbers)Verbal ExpressionDivisionThe quotient of 6 and a numberA number divided by 15half a number 7 ApplicationsCautionCAUTIONBecause subtraction and division are not commutative operations, be careful to correctly translate expressions involving them. For example, 5 less than a number is translated as x 5, not 5 x. A number subtracted from 12 is expressed as 12 x, not x 12. For division, the number by which we are dividing is the denominator, and the number into which we are dividing is the numerator. For example, a number divided by 15 and 15 divided into x both translate as . Similarly, the quotient of x and y is translated as . 8 Applications

Indicator Words for Equality

EqualityThe symbol for equality, , is often indicated by the word is. In fact, any words that indicate the idea of sameness translate to . 9ApplicationsTranslating Words into EquationsVerbal SentenceEquationTwice a number, decreased by 4, is 32.2x 4 3216x 25 87If the product of a number and 16 is decreased by 25, the result is 87. 48The quotient of a number and the number plus 6 is 48. x 54The quotient of a number and 8, plus the number, is 54. 10Applications Distinguishing between Expressions and EquationsDecide whether each is an expression or an equation.(a)4(6 x) 2x 1There is no equals sign, so this is an expression.(b)4(6 x) 2x 1 15Because of the equals sign, this is an equation.Note that the expression in part (a) simplifies to the expression 2x 23 and the equation in part (b) has solution 19. 116.6 Applications Involving Equations 12 ApplicationsSix Steps to Solving Application ProblemsSix Steps to Solving Application ProblemsStep 1 Read the problem, several times if necessary, until you understand what is given and what is to be found. Step 2 If possible draw a picture or diagram to help visualize the problem. Step 3 Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. Express any other unknown values in terms of the variable. Step 4 Write an equation using the variable expression(s). Step 5 Solve the equation. Step 6 Check the answer in the words of the original problem. 13 Applications Solving a Geometry ProblemThe length of a rectangle is 2 ft more than three times the width. The perimeter of the rectangle is 124 ft. Find the length and the width of the rectangle.Step 1Read the problem. We must find the length and width of the rectangle. The length is 2 ft more than three times the width and the perimeter is 124 ft.Step 2Assign a variable. Let W the width then 2 3W length. Make a sketch.W2 3WStep 3Write an equation. The perimeter of a rectangle is given by the formula P 2L 2W.Let L 2 3W and P 124.124 2(2 3W) 2W 14ApplicationsSolving a Geometry ProblemThe length of a rectangle is 2 ft more than three times the width. The perimeter of the rectangle is 124 ft. Find the length and the width of the rectangle.Step 4Solve the equation obtained in Step 3. 124 2(2 3W) 2W124 4 6W 2WRemove parentheses124 4 8WCombine like terms.124 4 4 8W 4Subtract 4.120 8W120 8WDivide by 8.

8 815 W 15ApplicationsSolving a Geometry ProblemThe length of a rectangle is 2 ft more than three times the width. The perimeter of the rectangle is 124 ft. Find the length and the width of the rectangle.Step 5State the answer. The width of the rectangle is 15 ft and the length is 2 3(15) 47 ft.Step 6Check the answer by substituting these dimensions into the words of the original problem. 16Saw a board 8 ft 4 in into nine equal pieces. If the loss per cut is 1/8 in, how long will each piece be?

Step 1 the board is to cut into 9 equal parts with 1/8 in wasted each cut. Since the measures are mixed ft and in convert to in.

Step 2 draw a picture.

100 in 17Saw a board 8 ft 4 in into nine equal pieces. If the loss per cut is 1/8 in, how long will each piece be?

Assign a variable for the unknown.

Let x the length of each equal piece.

Write an equation

100 inxxxxxxxxx 18Saw a board 8 ft 4 in into nine equal pieces. If the loss per cut is 1/8 in, how long will each piece be?

Solve the equation.

19Saw a board 8 ft 4 in into nine equal pieces. If the loss per cut is 1/8 in, how long will each piece be?

Each piece should be 11 in long.

Check in the problem.

20Distribute 1000 into 3 parts so that one part will three times as large as the second and the third part will be as large as the sum of the other two.

Read carefully.

Make a table

Assign a variable. Since there are 3 unknowns we need 2 more expressions using the variable .

Write an equation

x3x3x x 21Distribute 1000 into 3 parts so that one part will three times as large as the second and the third part will be as large as the sum of the other two.

Solve the equation

22Applications of Linear EquationsSolving an Investment ProblemA local company has 50,000 to invest. It will put part of the money in an account paying 3 interest and the remainder into stocks paying 5. If the total annual income from these investments will be 2180, how much will be invested in each account?Step 1Read the problem. We must find the amount invested in each account.Step 2Assign a variable. The formula for interest is I p r t.Let x the amount to invest at 3 50,000 x the amount to invest at 5.Rate (as a decimal)InterestPrincipleTime.030.03xx1.0550,000 x1.05(50,000 x)50,0002180? Totals ? 23Applications of Linear EquationsSolving an Investment ProblemA local company has 50,000 to invest. It will put part of the money in an account paying 3 interest and the remainder into stocks paying 5. If the total annual income from these investments will be 2180, how much will be invested in each account?Step 3Write an equation. The last column of the table gives the equation.interest at 3interest at 5

total interest

.03x.05(50,000 x)

2180

24Applications of Linear EquationsSolving an Investment ProblemA local company has 50,000 to invest. It will put part of the money in an account paying 3 interest and the remainder into stocks paying 5. If the total annual income from these investments will be 2180, how much will be invested in each account?Step 4Solve the equation. We do so without clearing decimals..03x .05(50,000) .05x 2180Distributive property.03x 2500 .05x 2180Multiply..02x 2500 2180Combine like terms..02x 320Subtract 2500x 16,000Divide by .02. 25Applications of Linear EquationsSolving an Investment ProblemA local company has 50,000 to invest. It will put part of the money in an account paying 3 interest and the remainder into stocks paying 5. If the total annual income from these investments will be 2180, how much will be invested in each account?Step 5State the answer. The company will invest 16,000 at 3. At 5, the company will invest 50,000 16,000 34,000.Step 6Check by finding the annual interest at each rate they should total 2180.and0.03(16,000) 480.05(34,000) 1700480 1700 2180, as required. 26Solving a Mixture ProblemA chemist must mix 12 L of a 30 acid solution with some 80 solution to get a 60 solution. How much of the 80 solution should be used?

EXAMPLE 7

Step 1Read the problem. The problem asks for the amount of 80 solution to be used.Step 2Assign a variable. Let x the number of liters of 80 solution to be used.80

30803012 LUnknown number of liters, x(12 x)L 27A chemist must mix 12 L of a 30 acid solution with some 80 solution to get a 60 solution. How much of the 80 solution should be used?

Write an equation.

308012 L x(12 x)L 28A chemist must mix 12 L of a 30 acid solution with some 80 solution to get a 60 solution. How much of the 80 solution should be used?

Solve the equation.

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